Answer to Question #153503 in Differential Equations for Ashweta Padhan

Question

Reduce the equation into canonical form and find the solution

sin²x u_xx+2cosx u_xy - u_yy=0

Answer 1

$sin^2x.u_{xx}+2cosx.u_{xy}-u_{yy}=0\;\;\;\;...(1)\\ A=sin^2x,\;B=2cosx,\;C=-1$

$B^2-4AC=4(cos^2x+sin^2x)=4>0$

So, (1) hyperbolic.

$A\lambda^2-B\lambda+C=0\\ \Rightarrow \lambda=cosx+1\;\;or, \;cosx-1\\\;\\ \therefore \frac{dy}{dx}=cosx+1\;or, \;cosx-1\\ \Rightarrow y=sinx+x+c_1\;\;or, \;sinx-x+c_2\\ let, \phi(x,y)=y-sinx-x ,\;\;\varphi(x,y)=y-sinx+x$

Let, $\zeta=\phi(x,y), \;\; \eta=\varphi (x,y)$

So,

$\zeta =\phi(x,y)=y-sinx-x\\ \zeta_x=-cosx-1\\ \zeta _y=1\\ \zeta_{yy}=0\\ \zeta _{xx}=sinx \\ and,\\ \eta =\varphi(x,y)=y-sin x+x\\ \eta _x=-cosx+1\\ \eta_y=1\\ \eta_{yy}=0\\ \eta_{xx}=sinx$

Now,

$u_{xx} = u_{\zeta}(\zeta_x^2) + 2u_{\zetaη}\zeta_xη_x + u_{ηη}η_x^2 + u_{\zeta}\zeta_{xx} + u_ηη_{xx}\\ u_{xy} = u_{\zeta\zeta}\zeta_x\zeta_y + u_{\zetaη}(\zeta_xη_y + \zeta_yη_x) + u_{ηη}η_xη_y + u_\zeta\zeta_{xy} + u_ηη_{xy}\\ u_{yy} = u_{\zeta\zeta}\zeta_y^2 + 2u_{\zetaη}\zeta_yη_y + u_{ηη}η_y^2 + u_\zeta\zeta_{yy} + u_ηη_{yy}$

Putting above values in $(1)$,

we get,

$\left(cos^4x+2cos^2x+3\right)u_{\zeta\eta}\\ +u_{\zeta\zeta}(cos^4x+2cos^3x+2cos^2x)\\+u_{\eta\eta}(cos^4x-2cos^3x+2cos^2x)=u_{\zeta}sin^3x\;\;\;...(2)\\$

From expressions of $\zeta \;and\;\eta,$

$x=\frac{1}{2}(\eta-\zeta)\\$

Now from (2),

$\left(cos^4\frac{1}{2}(\eta-\zeta)+2cos^2\frac{1}{2}(\eta-\zeta)+3\right)u_{\zeta\eta}\\ +u_{\zeta\zeta}(cos^4\frac{1}{2}(\eta-\zeta)+2cos^3\frac{1}{2}(\eta-\zeta)+2cos^2\frac{1}{2}(\eta-\zeta))\\+u_{\eta\eta}(cos^4\frac{1}{2}(\eta-\zeta)-2cos^3\frac{1}{2}(\eta-\zeta)+2cos^2\frac{1}{2}(\eta-\zeta))\\=u_{\zeta}sin^3\frac{1}{2}(\eta-\zeta)\;\;\;...(2)\\$

Hence the required canonical form is :

$$\left(cos^4\frac{1}{2}(\eta-\zeta)+2cos^2\frac{1}{2}(\eta-\zeta)+3\right)u_{\zeta\eta}\\ +u_{\zeta\zeta}(cos^4\frac{1}{2}(\eta-\zeta)+2cos^3\frac{1}{2}(\eta-\zeta)+2cos^2\frac{1}{2}(\eta-\zeta))\\+u_{\eta\eta}(cos^4\frac{1}{2}(\eta-\zeta)-2cos^3\frac{1}{2}(\eta-\zeta)+2cos^2\frac{1}{2}(\eta-\zeta))\\=u_{\zeta}sin^3\frac{1}{2}(\eta-\zeta)\;\;\;$$